How do you prove #cos(2x + π) = cos^2 (x - π/2) + cos(x + π) sin(x + π/2)# using the double angle identity? Trigonometry Trigonometric Identities and Equations Double Angle Identities 1 Answer P dilip_k Jul 8, 2016 #RHS=cos^2(x-pi/2)+cos(x+pi)sin(x+pi/2)# #=sin^2x-cosxcosx# #=-(cos^2x-sin^2x)# #=-cos2x# #=cos(pi+2x)=cos(2x+pi)=LHS# Answer link Related questions What are Double Angle Identities? How do you use a double angle identity to find the exact value of each expression? How do you use a double-angle identity to find the exact value of sin 120°? How do you use double angle identities to solve equations? How do you find all solutions for #sin 2x = cos x# for the interval #[0,2pi]#? How do you find all solutions for #4sinthetacostheta=sqrt(3)# for the interval #[0,2pi]#? How do you simplify #cosx(2sinx + cosx)-sin^2x#? If #tan x = 0.3#, then how do you find tan 2x? If #sin x= 5/3#, what is the sin 2x equal to? How do you prove #cos2A = 2cos^2 A - 1#? See all questions in Double Angle Identities Impact of this question 1662 views around the world You can reuse this answer Creative Commons License